The World of Absolute Inequalities

by on November 27th, 2012

Some GMAT problems present inequalities involving absolute value. Dealing with these questions is a multi-step process:

First, isolate the absolute value so that it stands alone on one side of the inequality.

Next, differentiate between two different cases: the number case, and the variable case.

1) The number case

In the number case, the other side is a number, i.e., the inequality has the form:

|something| < Number or |something| > Number

For example:

If |x-6|< 2, what is the range of values for x?

Solve absolute values of the number case by considering two possible scenarios:

First scenario – copy the inequality without the absolute value brackets and solve:

x-6 < 2  –> x < 2+6 = 8

Second scenario – remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign:

x-6 > -2″ title=”x-6 > -2″/> –> <img src=.

2) The variable case

If the inequality has an absolute value on one side, and variables on the other side, you cannot use the two-scenario approach. We call this the variable case, and it requires finding out what the inequality really means, rather than simply solving for x. Investigate the variable case by plugging in simple numbers for the variable(s) until you find a pattern.

In addition, remember that anything greater than an absolute value must be positive, regardless of the content of the absolute value. Many GMAT questions of the variable case depend on recognizing this simple concept for easy solution.

For example:

What is the range of x for which |x+1| < x?

This inequality does not fit the |something | <  Number form, so the two scenario approach is out. Instead, investigate the inequality – what can you learn? Which numbers can you plug in?

For x to be greater than an absolute value, x must be positive. Therefore, plug in simple positive numbers to find out when the inequality is true.

If x=1, then the inequality reads |1+1|<1, and is NOT true. So x cannot be 1.

If x=2, then the inequality reads |2+1|<2, and is NOT true. So x cannot be 2 either.

Can you find a pattern? In other words, do you think that the inequality would be true for x=3? or for x=0.5?

If x is positive, then x+1 is positive as well, and |x+1| = x+1. But then the inequality cannot be true because it reads x+1 < x. It follows that there is no value of x that fits in the inequality.

Remember:

Solve inequalities with an absolute value on one side and a number on the other by considering two scenarios:

  1. First scenario – copy the inequality without the absolute value brackets and solve.
  2. Second scenario – remove the absolute value brackets. Put a negative sign around the other side of the inequality, AND flip the sign. Solve.

When solving inequalities with an absolute value on one side and a variable on the other, the two scenario approach does not help. Instead:

  1. Investigate the variable case by plugging in simple numbers for the variable(s) until you find a pattern.
  2. Remember anything greater than an absolute value must be positive, regardless of the content of the absolute value. Many GMAT questions of the variable case depend on recognizing this simple concept for easy solution.

Now, try solving the following data sufficiency question:

Is m>n?

(1) |m+n| < |m| + |n|

(2) |m| > |n| + 1

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked;
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked;
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked;
D. EACH statement ALONE is sufficient to answer the question asked;
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

7 comments

  • The answer should be B

  • B isn;t right.  M=10 & N=5, (m>n)  M=-10 & N=-5, (m<n) both of these would satisfy statement B.

  • I think the answer is E.
    1. m and n are of different signs,but we cannot determine whether m>n
    2. From this, m and n could be of same sign or different sign, but cannot determine whether m>n
    m = -5, n = +3
    m = +5, n = -3
    m = +5, n = +3

    Combining 1 and 2, we only know that m and n are of different signs and absolute value of m is 1 more than absolute value of n, but still cannot determine whether m>n.

    I close my case here :)

  • Hi Prashant,
    Can you please elaborate a bit on ur analysis and answer.

    • Hi Nitin
      1. This inequality will be true only if m and n have different signs. You can take any value for m and n with different sign and plugging in the values.
      When m is -ve, then m n
      Hence we cannot determine whether m > n
      Insufficient

      2. In this case, absolute value of m is one more than absolute value of n..
      Lets pick few 
      a. m = -5, n = +3 
          |m| = |-5| = 5
          |n| + 1 = |3| + 1 = 4
          5 > 4
          m 4
         m > n

      c. m = +5, n = +3
          |m| = |5| = 5
          |n| + 1 = |3| + 1 = 4
          5 > 4
          m > n

      From Above three cases we can see that (2) is insufficient to prove m > n

      Combining (1) and (2): m and n are of different signs and absolute value of m and 1 more than absolute value of n.
      We see that (a) and (b) in the analysis of 2 satisfy both cases, but still we cannot determine whether m > n.

      Please let me know whether this makes sense.

      Regards,
      Prashant

    • Hi Nitin,
      Seems copy and paste did not work properly in my previous reply. I am reposting my reply.

      1. This inequality will be true only if m and n have different signs. You can take any value for m and n with different sign and plugging in the values.
      When m is -ve, then m n
      Hence we cannot determine whether m > n
      Insufficient

      2. In this case, absolute value of m is one more than absolute value of n..
      Lets pick few 
      a. m = -5, n = +3 
      |m| = |-5| = 5
      |n| + 1 = |3| + 1 = 4
      5 > 4
      m 4
      m > n

      c. m = +5, n = +3
      m| = |5| = 5
      |n| + 1 = |3| + 1 = 4
      5 > 4
      m > n

      From Above three cases we can see that (2) is insufficient to prove m > n

      Combining (1) and (2): m and n are of different signs and absolute value of m and 1 more than absolute value of n.
      We see that (a) and (b) in the analysis of 2 satisfy both cases, but still we cannot determine whether m > n.

  • Hi Nitin,
    Seems copy and paste did not work properly in my last two reples. I am reposting my reply, and hope this time it comes out ok. Sorry about this.

    1. This inequality will be true only if m and n have different signs. You can take any value for m and n with different sign and plugging in the values.
    When m is -ve, then m n
    Hence we cannot determine whether m > n
    Insufficient

    2. In this case, absolute value of m is one more than absolute value of n..
    Lets pick few numbers for m and n
    a. m = -5, n = +3 
    |m| = |-5| = 5
    |n| + 1 = |3| + 1 = 4
    5 > 4
    m 4
    m > n

    c. m = +5, n = +3
    m| = |5| = 5
    |n| + 1 = |3| + 1 = 4
    5 > 4
    m > n
    From Above three cases we can see that (2) is insufficient to prove m > n

    Combining (1) and (2): m and n are of different signs and absolute value of m is 1 more than absolute value of n.
    We see that (a) and (b) in the analysis of 2 satisfy both cases, but still we cannot determine whether m > n.

    Hope this makes sense. 
    Regards,
    Prashant

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